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A single slit diffraction pattern is formed with white light. For what wavelength of light, the $3^{\text {rd }}$ secondary maximum in diffraction pattern coincides with the $2^{\text {nd }}$ secondary maximum in the pattern of red light of wavelength $6000 Å$ ?
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$4300 Å$
$\begin{aligned} & \text { Distance of nth maximum } \mathrm{y}_{\mathrm{n}}=\left(\mathrm{n}+\frac{1}{2}\right) \frac{\lambda \mathrm{D}}{\mathrm{a}} \\ & \therefore \mathrm{y}_2=\left(2+\frac{1}{2}\right) \frac{\lambda_1 \mathrm{D}}{\mathrm{a}} \text { and } \mathrm{y}_3=\left(3+\frac{1}{2}\right) \frac{\lambda_2 \mathrm{D}}{\mathrm{a}} \\ & \therefore 2.5 \lambda_1=3.5 \lambda_2 \\ & \therefore 2.5 \times 6000=3.5 \lambda_2 \\ & \therefore \lambda_2=\frac{2.5 \times 6000}{3.5} \simeq 4300 A\end{aligned}$
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